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I am having some problems to understand the meaning of the following theorem due to Roitmann. I found this theorem in Voisin's book: Hodge Theory and Complex Algebraic Geometry, Volume II, page 289.

Theorem 10.14. The Albanese map \begin{equation*} alb_X:CH_0(X)_{hom}\rightarrow Alb(X) \end{equation*} Induces an isomorphism on torsion points.

What does it mean that the isomorphism is on torsion points? what is an intuitive way to think about torsion points?

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    $\begingroup$ It means that the subgroups on the left and right consisting of torsion points are isomorphic. $\endgroup$
    – Donu Arapura
    Commented Apr 13, 2022 at 9:04
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    $\begingroup$ Perhaps the confusion is that on the left we have a group on the right a group variety. The term "torsion points" refers to points in the group variety $Alb(X)$ which are torsion. On the left hand side this is just torsion elements of the group. $\endgroup$
    – Kapil
    Commented Apr 13, 2022 at 9:17

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It occurs to me that your question shouldn't be taken literally, and is simply asking about the meaning of the theorem. To appreciate it, one can ask what $CH_0(X)$ looks like. As a first attempt, map it to something more concrete like the Albanese (which is just a complex torus over $\mathbb{C}$). It is easy to see that the map is surjective, and an isomorphism for curves. In higher dimensions, Mumford showed that the kernel can be huge. Given this, Roitman's theorem that the Albanese map restricted to the torsion subgroups yields an isomorphism is pretty surprising. For example if $H_1(X,\mathbb{Z})=0$, the theorem would imply that the torsion subgroup of $CH_0(X)$ is trivial.

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